Algebra Know-It-ALL

Part Two 333

along the line until y= 0, we’ll end up on the x axis, and we’ll have gone 8/2, or 4, units in the
negative x direction from the y axis. That means the x-intercept is −4. It looks that way in the
figure; now we know it’s really true.

Question 17-3

How can we verify that the preceding answer is correct by manipulating the equation for the
line?

Answer 17-3

We can start with either the original equation or the SI form with y as the dependent variable,
and manipulate things until we get the SI form with x all alone on the left side of the equals
sign. In Answer 16-4, that was done starting with the original equation. If we start with the SI
form with y as the dependent variable, we can proceed like this:

y= 2 x+ 8

y− 8 = 2 x

(1/2)y− 4 =x

x= (1/2)y− 4

This SI equation tells us that the x-intercept is equal to −4.

Question 17-4

How can we determine the x-intercept of the line representing the equation y= (1/3)x+ 3 on
the basis of the known slopes and the point data in Fig. 20-6? (As in Question 17-2, we can
sety= 0 and then solve for x; but again, this exercise is meant to show how slope and intercept
are related geometrically.)

Answer 17-4

The slope of the line is 1/3. Therefore, if we start from any point on the line and move in
the positive x direction by n units, we must move in the positive y direction by n/3 units to
stay on the line. In the opposite sense, if we start from any point on the line and move in the
negative y direction by p units, we must move in the negative x direction by 3p units to stay
on the line. Let’s start at the point (0, 3), which is the y-intercept. If we move in the negative
y direction along the line until y= 0, we’ll end up on the x axis, and we’ll have gone 3 × 3, or
9, units in the negative x direction from the y axis. That means the x-intercept is −9. This is
outside the field of view in Fig. 20-6.

Question 17-5

How can we verify that the preceding answer is correct by manipulating the equation for the line?

Answer 17-5

As we did in Answer 17-3, we can start with either the original equation or the SI form with y as
the dependent variable, and manipulate things until we get the SI form with x all alone on the